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For n >= 1, a(n) tells the one-based position of the digit (from the right) where the iteration stopped at, when constructing a Semigreedy Catalan representation of n as described in A244159.

Algorithm for constructing the sequence: Find the largest Catalan number which is less than or equal to n (this is A081290(n) = A000108(k), where k = A244160(n), that is, the corresponding index of that Catalan number), and subtract that from n. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the remaining n, and if it does not, then subtract that also from n, and keep on doing the same for lesser and lesser Catalan numbers, comparing and also subtracting them (whenever it is possible without going less than zero) from n, until either n becomes zero, or after subtracting C(1) = 1 from n, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to remaining n, and start the process again. However, when at some point n finally reaches zero, then the index k of the last Catalan number, A000108(k) which was subtracted from n before it reached zero, is our result, a(n) = k. [Here n = the original value of n, from which we started subtracting initially from].

If n is one of the terms of A197433, meaning that if it can be represented as a sum of distinct Catalan numbers as n = C(i) + C(j) + ... + C(k) (which representation then necessarily is unique), then a(n) = min(i,j,...,k).

For n >= 1, a(n) tells the zero-based position of the digit (from the right) where the iteration stopped at, when constructing a Semigreedy Catalan representation of n as described in A244159.

Algorithm for constructing the sequence: Define a(0) as 0, and for larger values of n, find first the largest Catalan number which is less than or equal to n [which is A081290(n)], and the index k = A244160(n), of that Catalan number. Initialize a vector of k zeros, [0, 0, ..., 0]. Set n_remaining = n - A000108(k) and add 1 to the leftmost element of vector, so that it will become [1, 0, ..., 0]. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the n_remaining, and provided that C(m) <= n_remaining, then set n_remaining = n_remaining - C(m) and increment by one the m-th element of the vector (where the 1st element is the rightmost), otherwise just decrement m by one and keep on doing the same with lesser and lesser Catalan numbers, and whenever it is possible to subtract them from n_remaining (without going less than zero), do so and increment the corresponding m-th element of the vector, as long as either n_remaining becomes zero, or after subtracting C(1) = 1 from n_remaining, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to n_remaining, and start the process again. However, after a finite number of such iterations, n_remaining will finally reach zero, and the result of a(n) is then the vector of numbers constructed, concatenated together and represented as a decimal number.

This shares with "Greedy Catalan Base" (A014418) the property that a simple weighted sum of Sum_{k=1..} digit(k)*C(k) recovers the natural number n, which the given numeral string like A014418(n) or here, a(n), represents. (Here C(k) = the k-th Catalan number, A000108(k), and digit(1) = the digit in the rightmost, least significant digit position.)

In this case, A244158(a(n)) = n holds for only up to 33603, after which comes the first representation containing a "digit" larger than nine, at a(33604), where the underlying string of numbers is [1,2,3,4,5,6,7,8,9,10] but the decimal system used here can no more unambiguously represent them.

On the other hand, with the given Scheme-functions, we always get n back with: (CatBaseSumVec (A244159raw n)).

For n >= 1, A014138(n) gives the positions of repunits: 1, 11, 111, 1111, ...

The "rep-2's": 22222, 222222, 2222222, 22222222, 222222222, ..., etc., occur in positions 128, 392, 1250, 4110, 13834, ... i.e. 2*A014138(n) for n >= 5.

Replace 2^k with A000108(k+1) in binary expansion of n.

From Antti Karttunen, Jun 22 2014: (Start)

On the other hand, A244158 is similar, but replaces 10^k with A000108(k+1) in decimal expansion of n.

This sequence gives all k such that A014418(k) = A239903(k), which are precisely all nonnegative integers k whose representations in those two number systems contain no digits larger than 1. From this also follows that this is a subsequence of A244155.

(End)

a(n) <= 1;

a(A000108(n)) = 1; a(A141351(n)) = 1; a(A014138(n)) = 1.

A197433 gives all such numbers k that a(k) = 1, in other words, this is the characteristic function of A197433, and all three sequences mentioned above are its subsequences. - Antti Karttunen, Jun 25 2014

Note that a(33604) = A000217(10) = 55 because the sum is computed from the underlying list (vector) of numbers, and thus is not subject to any corruption by decimal representation as A244159 itself is.

Equivalent description: partition n "greedily" as terms of A197433, i.e. n = A197433(i) + A197433(j) + ... + A197433(k), always using the largest term of A197433 that still "fits in" (i.e. is <= n remaining). Then a(n) = A000120(i) + A000120(j) + ... + A000120(k).